Don't find all the number

We can use formula ,

$\frac{a}{b}×(sum of digits)×(111..._{‘b' times})$

Where:

$a$ =Number of numbers that can be formed.

$b$ =number of digits

Thus by calculating,

$\frac{5!}{5}×15×11111$ = $\boxed{3999960}$

Assume that the first number is $1$ ,

the second number has $4$ probabilities ( $5, 4, 3$ and $2$ ), then

the third number has $3$ probability left, then

the forth number has $2$ probability left, then

the last number has $1$ probability left.

$\Rightarrow$ the numbers of five digits numbers that starts with $1$ is $4*3*2*1=24$ . (Similarly, we can prove that the numbers of five digits numbers that starts with $2, 3, 4$ and $5$ is also $24$ .)

By the same way, we can prove that the number $1$ appears at other digits $24$ times. (And also the other numbers.)

When we add all the five digit numbers that can be formed using the digits $1,2,3,4$ and $5$ , every digits is added $24$ times.

Therefore, the sum of all five digit numbers that can be formed using the digits $1,2,3,4$ and $5$ (repetition of digits is not included)is

$24*(1+2+3+4+5)*11111\\=24*15*11111\\=360*11111\\=\boxed{3999960}$ .

Imagine writing all the possible numbers in a list like

12345

12354

12435

...

There are exactly 4! numbers in which the digit 1 is in the first place, and 4! as well where the digit is in 2. place and so on. Same goes for the other digits. Therefore adding all these numbers place by place we get $(1+2+3+4+5)*4!+(1+2+3+4+5)*4!*10+(1+2+3+4+5)*4!*10^2+(1+2+3+4+5)*4!*10^3+(1+2+3+4+5)*4!*10^4=3999960$

24×15×11111